Repost of this Math.SE question due to a lack of answers (No one was able to help me find the closed form of $U_T$ and $|T|$ after two bounties). I also searched extensively online but couldn't find anything. A reference for the solution suffices.

Define the Volterra integral operator (or numerical differentiation operator) $$ T: L^2([0,1]) \to L^2([0,1]), \ f(x) \mapsto \int_{0}^{x} f(y) dy. $$ Find its polar decomposition $T = U_T | T |$.

First I tried to find $|T| := (T^* T)^{\frac{1}{2}}$. We have $$ T^*: L^2([0,1]) \to L^2([0,1]), \ f(x) \mapsto \int_{x}^{1} f(y) dy. $$ and therefore $$ T^* T: L^2([0,1]) \to L^2([0,1]), \ f(x) \mapsto \int_{x}^{1} \int_{0}^{y} f(z) dz dy. $$ Now I got stuck when finding $\sqrt{T^* T}$. I found out that $T^2 \ne T^* T$ and $(T^*)^2 \ne T^* T$. Can somebody please give me a hint?

I have also tried writing out $T^* T$ more explicitly: denote by $F$ the anti-derivative of $f$ and by $\mathscr{F}$ the antiderivative of $F$. The we have $$ (T^* T f)(x) = \int_{x}^{1} F(y) - F(0) dy = \int_{x}^{1} F(y) dy - (1 - x) F(0) = \mathscr{F}(1) - \mathscr{F}(x) - (1 - x)F(0). $$

Edit. I also know that $$ (T^{n + 1} f) = \frac{1}{n!} \int_0^x (x - y)^n f(y) dy. $$ Maybe this can help use find a closed form for $(I - T^* T)^n$?

Edit 2. I know that $$ | T | = \sum_{n \in \mathbb N} \sigma_n \langle \cdot, v_n \rangle v_n, $$ where $\sigma_n := \frac{1}{\pi\left(n - \frac{1}{2}\right)}$ are the singular values of $T^* T$ and $(v_n)_{n \in \mathbb N} \subset L^2([0,1])$ a orthonormal basis of $\overline{\text{ran}(T)}$ which fulfils $$ T^* T x = \sum_{n \in \mathbb N} \sigma_n^2 \langle x, v_n \rangle, v_n. $$ The eigenfunctions of $T^* T$ are $f_n(x) = C \cdot \cos(\sigma_n^{-1} x)$ for some $C \in \mathbb{R}$.